SO SÁNH : A = 3^123 +1 / 3^125 + 1 VÀ B = 3^122 + 1 / 3^124 + 1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\dfrac{3^{123}+1}{3^{125}+1}\) Vì 3123 + 1 < 2125 + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)
A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B
Vậy A < B
Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\)N*)
Ta có:
\(A=\frac{3^{123}+1}{3^{125}+1}< \frac{3^{123}+1+2}{3^{125}+1+2}\)
\(A< \frac{3^{123}+3}{3^{125}+3}\)
\(A< \frac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\)
\(A< \frac{3^{122}+1}{3^{124}+1}=B\)
=> A < B
\(B=\frac{3^{122}}{3^{124}+1}=\frac{3^{123}}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+1}=A\)
Do đó \(A>B\).
\(A=\dfrac{3^{123}+1}{3^{125}+1}\Leftrightarrow3^2A=\dfrac{3^{125}+9}{3^{125}+1}\)
\(9A=\dfrac{3^{125}+1}{3^{125}+1}+\dfrac{8}{3^{125}+1}=1+\dfrac{8}{3^{125}+1}\)
\(B=\dfrac{3^{122}+1}{3^{124}+1}\Leftrightarrow3^2B=\dfrac{3^{124}+9}{3^{124}+1}\)
\(9B=\dfrac{3^{124}+1+8}{3^{124}+1}+\dfrac{3^{124}+1}{3^{124}+1}+\dfrac{8}{3^{124}+1}=1+\dfrac{8}{3^{124}+1}\)
\(9A< 9B\Leftrightarrow A< B\)
ai trả lời đúng mik tick cho